Question
Question: A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A sma...
A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle θ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is μ. Find the range of the angular speed for which the block will not slip.

Range of angular speed is [Rsinθg1+μtanθtanθ−μ,Rsinθg1−μtanθtanθ+μ] with appropriate conditions for μ and θ.
Solution
The problem involves finding the range of angular speeds for which a block in a rotating hemispherical bowl does not slip. This is determined by analyzing the forces acting on the block at the minimum and maximum angular speeds, considering static friction.
Case 1: Minimum Angular Speed (ωmin)
At this speed, the block tends to slip downwards, so friction acts upwards. We resolve forces into horizontal and vertical components to find the normal force N and then ωmin.
Vertical Equilibrium: Ncosθ+fssinθ−mg=0
Horizontal Motion (Centripetal Force): Nsinθ−fscosθ=mωmin2Rsinθ
Solving these equations, we find: ωmin=Rsinθg1+μtanθtanθ−μ (for tanθ≥μ)
If tanθ<μ, ωmin=0.
Case 2: Maximum Angular Speed (ωmax)
At this speed, the block tends to slip upwards, so friction acts downwards. We again resolve forces into horizontal and vertical components.
Vertical Equilibrium: Ncosθ−fssinθ−mg=0
Horizontal Motion (Centripetal Force): Nsinθ+fscosθ=mωmax2Rsinθ
Solving these equations, we find: ωmax=Rsinθg1−μtanθtanθ+μ (for tanθ<1/μ)
If tanθ≥1/μ, ωmax=∞.
Therefore, the range of angular speed is [ωmin,ωmax].